2007 Alabama ARML TST Problems/Problem 14
Contents
[hide]Problem
Find the sum of the real roots of
Solution
Step 1: using the symmetry
Let
The polynomial is symmetric. Each symmetric polynomial has the following property: for
,
is a root if and only if
is.
To see why this is true, assume that we have a such that
. We can now divide both sides by
. We get:
But the left hand side is just . Thus also
.
Step 2: multiplying three quadratic terms
Our polynomial has six complex roots: ,
,
,
,
, and
. It can be expressed as:
Let ,
, and
. We can then write:
We can now compare the coefficients of ,
and
in this expression. (Comparing other coefficients is not necessary due to symmetry.) We get the following equations:
Simplifying, we get:
Using Vieta's formulas, this means that ,
and
are the roots of the polynomial
.
Step 3: Finding a root of q
Obviously, is increasing for
, and
. Thus
has exactly one positive root. We can easily find that
. Thus WLOG
.
(Note:
Remembering that and
, we now have
and
.
From the first equation, let
. Substituting into the second one, we get
. This obviously has no real solution, thus
and
are not real. But we don't need this observation.)
Step 4: Partially dividing p
We now know that is a factor of
. We can partially divide
to get:
Step 5: Handling the remaining part of p
The polynomial has no real roots.
To prove this, one can for example write:
Step 6: Summary
We found that , where
has no real roots. Thus all real roots of
are the two real roots of
, and their sum is obviously
.
Note
roots of p computed by Octave: -4.48006 + 14.30270i -4.48006 - 14.30270i 8.88748 + 0.00000i 0.11252 + 0.00000i -0.01994 + 0.06367i -0.01994 - 0.06367i
See also
2007 Alabama ARML TST (Problems) | ||
Preceded by: Problem 13 |
Followed by: Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |